Question

The probability distribution of a random variable X is given below:

X	0	1	2	3
P(X)	k	$\frac{k}{2}$	$\frac{k}{4}$	$\frac{k}{8}$

(i) Determine the value of k

(ii) Determine P(X ≤ 2) and P(X > 2)

(iii) Find P(X ≤ 2) + P(X > 2)

Answer 1

We have,

The probability distribution of a random variable X is given below:

X	0	1	2	3
P(X)	k	$\frac{k}{2}$	$\frac{k}{4}$	$\frac{k}{8}$

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{As},\sum p_{\mathit{i}}=1 \\ \Rightarrow k+\frac{k}{2}+\frac{k}{4}+\frac{k}{8}=1 \\ \Rightarrow \frac{8k+4k+2k+k}{8}=1 \\ \Rightarrow \frac{15k}{8}=1 \\ \therefore k=\frac{8}{15} \\ \left(\mathrm{ii}\right)\mathrm{As},\mathrm{P}\left(X\le 2\right)=1-\mathrm{P}\left(X=3\right) \\ =1-\frac{k}{8} \\ =1-\frac{8}{15\times 8} \\ =1-\frac{1}{15} \\ =\frac{14}{15} \\ \mathrm{Also},\mathrm{P}\left(X>2\right)=\mathrm{P}\left(X=3\right) \\ =\frac{8}{15\times 8} \\ =\frac{1}{15} \\ \left(\mathrm{iii}\right)\mathrm{P}\left(X\le 2\right)+\mathrm{P}\left(X>2\right) \\ =\frac{14}{15}+\frac{1}{15}\left[\mathrm{Using}\left(\mathrm{ii}\right)\right] \\ =\frac{15}{15} \\ =1 \end{gathered}$$